What is the integral of 1/(usqrt(1+u^2)) ?

The integral of 1/(usqrt(1+u^2)) is -1/2 ln|sqrt(1+u^2)+1|+1/2 ln|sqrt(1+u^2)-1|+C

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integral of 1/(usqrt(1+u^2))

\int \frac{1}{u 1 + u ^{2}} d u

- \frac{1}{2} ln 1 + u^{2} + 1 + \frac{1}{2} ln 1 + u^{2} - 1 + C

Solution steps

\int \frac{1}{u 1 + u ^{2}} d u

Apply u-substitution

= \int \frac{1}{v ^{2} - 1} d v

Factor

v^{2} - 1 : - (- v^{2} + 1)

= \int \frac{1}{- ( - v ^{2} + 1 )} d v

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= - \int \frac{1}{- v ^{2} + 1} d v

Use the common integral:

\int \frac{1}{- v ^{2} + 1} d v = \frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2}

= - (\frac{ln ∣ v + 1 ∣}{2} - \frac{ln ∣ v - 1 ∣}{2})

Substitute back

v = 1 + u^{2}

= - (\frac{ln 1 + u ^{2} + 1}{2} - \frac{ln 1 + u ^{2} - 1}{2})

Simplify

- (\frac{ln 1 + u ^{2} + 1}{2} - \frac{ln 1 + u ^{2} - 1}{2}) : - \frac{1}{2} ln 1 + u^{2} + 1 + \frac{1}{2} ln 1 + u^{2} - 1

= - \frac{1}{2} ln 1 + u^{2} + 1 + \frac{1}{2} ln 1 + u^{2} - 1

Add a constant to the solution

= - \frac{1}{2} ln 1 + u^{2} + 1 + \frac{1}{2} ln 1 + u^{2} - 1 + C

\frac{d}{d x} (7 sin (337 x + \frac{7 π}{18}))

\int \frac{2 x + 6}{x ^{2} + 2 x + 5} d x

p a r t ia l f r a c t i o n \frac{x ^{2}}{1 - x ^{2}}

(y^{3} - 1) e^{x} + 3 y^{2} (e^{x} + 1) \frac{d y}{d x} = 0, y (0) = 2

t a y l or 5^{t}, 0

What is the integral of 1/(usqrt(1+u^2)) ?
The integral of 1/(usqrt(1+u^2)) is -1/2 ln|sqrt(1+u^2)+1|+1/2 ln|sqrt(1+u^2)-1|+C